If x and y are different integers and x^2 = xy, which of the

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If x and y are different integers and x^2 = xy, which of the [#permalink]

15 Jul 2007, 08:15

If x and y are different integers and x^2 = xy, which of the following must be true ?

1. x = 0
2. y = 0
3. x = -y

(a). I only
(b). II only
(c). III only
(d). I and III only
(e). I, II, and III

Please provide explaination...Thanks
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Re: different integers [#permalink]

15 Jul 2007, 08:37

alimad wrote:

A for me.

Re-arrange equation: x^2 - xy = 0 => x*(x-y) = 0
Therefore, either x=0 or x=y. We know that x and y are different, so x must be equal to 0.

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[#permalink]

15 Jul 2007, 08:41

Answer choice A is correct.

The question gives you x^2=xy
you can simplify that as x^2-xy=0
x(x-y)=0 the solution will be either x=0 or x=y, since x and y are different integers x cannot equal y, that leaves you x=0 to be the only answer.

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Re: different integers [#permalink]

15 Jul 2007, 10:45

alimad wrote:

x^2 - xy = 0; x=0 or x=y but it is provided that x<>y
so only A.

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16 Jul 2007, 10:00

I have a question. 4 and -4 are the same integer?

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Re: different integers [#permalink]

07 Oct 2011, 17:33

Doesn't x=-1 y=1 disprove A?

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Re: different integers [#permalink]

07 Oct 2011, 17:48

ekang1026 wrote:

Doesn't x=-1 y=1 disprove A?
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x=-1 AND y=1 doesn't satisfy the expression x^2=xy

x^2=xy
(-1)^2=-1*1
1=-1. Not correct.

x^2=xy
x^2-xy=0
x(x-y)=0
Means;
x=y OR x=0
We know x \ne y. So, x must be equal to 0.
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Re: different integers [#permalink]

03 Nov 2011, 15:23

Okay, something is off here for me

What I did is this

X ^ 2 = xy

I took the square root of x ^ 2 so I ended up with

x = square root of (x.y)
which means x = square root of (x) * square root of (y)

Therefore, square root of (y) will equal x/square root of (x)

If x is = 0 then it will be undefined. You can't divide by 0.

The same applies to the value of square root of (x) which will equal x/square root of (y). Therefore, y can't equal 0.

Anyone help me here?

Thanks

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Re: different integers [#permalink]

03 Nov 2011, 23:07

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Lstadt wrote:

First of all, square root of x^2 is |x|, not x. Do not take the square root until and unless you really need to.
x^2 = xy
You do not divide both sides by x here. You lose out on a solution.
What you can very safely do is x^2 - xy = 0
x(x-y) = 0
Now, either x = 0 or x = y or both
Since x and y are different, x must be 0.

When x = 0, no matter what the value of y, x^2 is equal to xy since both sides are equal to 0.
Since x = 0, you cannot re-write this as y = x^2/x
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Re: different integers [#permalink] 03 Nov 2011, 23:07

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